5w^2-4w-224=0

Solution for 5w^2-4w-224=0 equation:

5w^2-4w-224=0

a = 5; b = -4; c = -224;
Δ = b2-4ac
Δ = -42-4·5·(-224)
Δ = 4496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4496}=\sqrt{16*281}=\sqrt{16}*\sqrt{281}=4\sqrt{281}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{281}}{2*5}=\frac{4-4\sqrt{281}}{10}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{281}}{2*5}=\frac{4+4\sqrt{281}}{10}
$